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Intel 8086MICROPROCESSOR

Intel 8086MICROPROCESSOR

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Features Features

• It is a 16-bit μp.

• 8086 has a 20 bit address bus can access up to 220 memory locations (1 MB).

• It has multiplexed address and data bus AD0- AD15 and A16 – A19.

• It requires +5V power supply.

• A 40 pin dual in line package.

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• 8086 is designed to operate in two modes, Minimum and Maximum.

• It can prefetches up to 6 instruction bytes from memory and queues them in order to speed up instruction execution.

• Address ranges from 00000H to FFFFFH

• Memory is byte addressable - Every byte has a separate address.

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Intel 8086 Internal ArchitectureIntel 8086 Internal Architecture

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Internal architecture of 8086Internal architecture of 8086

• 8086 has two blocks BIU and EU.

• The BIU handles all transactions of data and addresses on the buses for EU.

• The BIU performs all bus operations such as instruction fetching, reading and writing operands for memory and calculating the addresses of the memory operands. The instruction bytes are transferred to the instruction queue.

• EU executes instructions from the instruction system byte queue.

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• Both units operate asynchronously to give the 8086 an overlapping instruction fetch and execution mechanism which is called as Pipelining. This results in efficient use of the system bus and system performance.

• BIU contains Instruction queue, Segment registers, Instruction pointer, Address adder.

• EU contains Control circuitry, Instruction decoder, ALU, Pointer and Index register, Flag register.

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BUS INTERFACE UNIT (BIU)BUS INTERFACE UNIT (BIU)Contains• 6-byte Instruction Queue (Q)• The Segment Registers (CS, DS, ES, SS).• The Instruction Pointer (IP).• The Address Summing block (Σ)

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THE QUEUE (Q)THE QUEUE (Q)

• The BIU uses a mechanism known as an instruction stream queue to implement a pipeline architecture.

• This queue permits pre-fetch of up to 6 bytes of instruction code. Whenever the queue of the BIU is not full, it has room for at least two more bytes and at the same time the EU is not requesting it to read or write operands from memory, the BIU is free to look ahead in the program by pre-fetching the next sequential instruction.

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• These pre-fetching instructions are held in its FIFO queue. With its 16 bit data bus, the BIU fetches two instruction bytes in a single memory cycle.

• After a byte is loaded at the input end of the queue, it automatically shifts up through the FIFO to the empty location nearest the output.

• The EU accesses the queue from the output end. It reads one instruction byte after the other from the output of the queue.

• The intervals of no bus activity, which may occur between bus cycles are known as Idle state.

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Opcode

Data

Opcode queue

From memory

Repeat the same procedure for successive contents of Q

Execute it with data bytes decoded by the decoder

Take 2nd byte from Q as opcode, decode 2nd byte opcode

Execute it with data bytes decoded by the decoder

Is it Single byte?

Take 2nd byte from Q as opcode, decode 2nd byte opcode

(Decode also decides the no. of data bytes for the instructions)

Update queue

Opcode 2nd byte

yes

No

The Queue Operation The Queue Operation

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Segmented MemorySegmented Memory

Code segment (64KB)

Data segment (64KB)

Extra segment (64KB)

Stack segment (64KB)

1 MB

The memory in an 8086/88 based system is organized as segmented memory.

The CPU 8086 is able to address 1Mbyte of memory.

The Complete physically available memory may be divided into a number of logical segments.

00000

FFFFF

Physical Memory

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• The size of each segment is 64 KB• A segment is an area that begins at any location which is

divisible by 16.• A segment may be located any where in the memory• Each of these segments can be used for a specific

function.

– Code segment is used for storing the instructions.– The stack segment is used as a stack and it is used to store the

return addresses.– The data and extra segments are used for storing data byte.

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• The 4 segments are Code, Data, Extra and Stack segments.• A Segment is a 64kbyte block of memory.• The 16 bit contents of the segment registers in the BIU

actually point to the starting location of a particular segment.• Segments may be overlapped or non-overlapped

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Segment registersSegment registers

• In 8086/88 the processors have 4 segments registers

• Code Segment register (CS), Data Segment register (DS), Extra Segment register (ES) and Stack Segment (SS) register.

• All are 16 bit registers.

• Each of the Segment registers store the upper 16 bit address of the starting address of the corresponding segments.

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Instruction pointer & summing blockInstruction pointer & summing block

• The instruction pointer register contains a 16-bit offset address of instruction that is to be executed next.

• The IP always references the Code segment register (CS).

• The value contained in the instruction pointer is called as an offset because this value must be added to the base address of the code segment, which is available in the CS register to find the 20-bit physical address.

• The value of the instruction pointer is incremented after executing every instruction.

• To form a 20bit address of the next instruction, the 16 bit address of the IP is added (by the address summing block) to the address contained in the CS , which has been shifted four bits to the left.

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• The following examples shows the CS:IP scheme of address formation:

Inserting a hexadecimal 0H (0000B)

with the CSR or shifting the CSR

four binary digits left

3 4 B A 0 ( C S ) +8 A B 4 ( I P )

3 D 6 5 4 (next address)

34BA 8AB4CS IP

34BA0

3D645

44B9F

Code segment

8AB4 (offset)

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• Example For Address Calculation (segment: offset)

• If the data segment starts at location 1000h and a data reference contains the address 29h where is the actual data?

Required Address

Offset

Segment Address

0000 0000 0010 1001

0000

0001 0000 0000 0010 1001

0001 0000 0000 0000

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EXECUTION UNITEXECUTION UNIT

• Decodes instructions fetched by the BIU• Generate control signals,• Executes instructions.

The main parts are:

• Control Circuitry• Instruction decoder• ALU

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AH AL

BH BL

CH CL

DH DL

SP

BP

SI

DI

8 bits 8 bits

16 bits

Accumulator

Base

Count

Data

Stack Pointer

Base Pointer

Source Index

Destination Index

AX

BX

CX

DX

Pointer

Index

8 bits 8 bits

16 bits

Accumulator

Base

Count

Data

Stack Pointer

Base Pointer

Source Index

Destination Index

EXECUTION UNIT – General Purpose RegistersEXECUTION UNIT – General Purpose Registers

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EXECUTION UNIT – General Purpose RegistersEXECUTION UNIT – General Purpose Registers

Register PurposeAX Word multiply, word divide, word I /O

AL Byte multiply, byte divide, byte I/O, decimal arithmetic

AH Byte multiply, byte divide

BX Store address information

CX String operation, loops

CL Variable shift and rotate

DX Word multiply, word divide, indirect I/O(Used to hold I/O address during I/O instructions. If the result is more than 16-bits, the lower order 16-bits are stored in accumulator and higher order 16-bits are stored in DX register)

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Pointer And Index RegistersPointer And Index Registers• used to keep offset addresses.

• Used in various forms of memory addressing.

• In the case of SP and BP the default reference to form a physical address is the Stack Segment (SS-will be discussed under the BIU)

• The index registers (SI & DI) and the BX generally default to the Data segment register (DS).

SP: Stack pointer– Used with SS to access the stack segment

BP: Base Pointer– Primarily used to access data on the stack– Can be used to access data in other segments

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• SI: Source Index register– is required for some string operations– When string operations are performed, the SI register points to memory locations in the data segment which is addressed by the DS register. Thus, SI is associated

with the DS in string operations.

• DI: Destination Index register– is also required for some string operations.– When string operations are performed, the DI register points to memory locations in the data segment which is addressed by the ES register. Thus, DI is associated

with the ES in string operations.

• The SI and the DI registers may also be used to access data stored in arrays

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EXECUTION UNIT – Flag RegisterEXECUTION UNIT – Flag Register

• A flag is a flip flop which indicates some conditions produced by the execution of an instruction or controls certain operations of the EU .

• In 8086 The EU contains

a 16 bit flag register

9 of the 16 are active flags and remaining 7 are undefined.

6 flags indicates some conditions- status flags

3 flags –control Flags

U U U U OF

DF

IF TF

SF

ZF

U AF

U PF

U CF

Carry Over flow Direction

Interrupt Trap

SignZero

AuxiliaryParity

U - Unused

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EXECUTION UNIT – Flag RegisterEXECUTION UNIT – Flag Register

Flag PurposeCarry (CF) Holds the carry after addition or the borrow after subtraction.

Also indicates some error conditions, as dictated by some

programs and procedures .

Parity (PF) PF=0;odd parity, PF=1;even parity.

Auxiliary (AF) Holds the carry (half – carry) after addition or borrow after

subtraction between bit positions 3 and 4 of the result

(for example, in BCD addition or subtraction.)

Zero (ZF) Shows the result of the arithmetic or logic operation.

Z=1; result is zero. Z=0; The result is 0

Sign (SF) Holds the sign of the result after an arithmetic/logic instruction

execution. S=1; negative, S=0

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Flag Purpose

Trap (TF)

A control flag.

Enables the trapping through an on-chip debugging

feature.

Interrupt (IF)

A control flag.

Controls the operation of the INTR (interrupt request)

I=0; INTR pin disabled. I=1; INTR pin enabled.

Direction (DF)

A control flag.

It selects either the increment or decrement mode for DI

and /or SI registers during the string instructions.

Overflow (OF)

Overflow occurs when signed numbers are added or

subtracted. An overflow indicates the result has exceeded

the capacity of the Machine

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Execution unit – Flag RegisterExecution unit – Flag Register

• Six of the flags are status indicators reflecting properties of the last arithmetic or logical instruction.

• For example, if register AL = 7Fh and the instruction ADD AL,1 is executed then the following happen

AL = 80hCF = 0; there is no carry out of bit 7PF = 0; 80h has an odd number of onesAF = 1; there is a carry out of bit 3 into bit 4ZF = 0; the result is not zeroSF = 1; bit seven is oneOF = 1; the sign bit has changed